Respuesta :
Answer:
0.4 = 40% probability that one die resulted in a 3.
Step-by-step explanation:
Conditional Probability
We use the conditional probability formula to solve this question. It is
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]
In which
P(B|A) is the probability of event B happening, given that A happened.
[tex]P(A \cap B)[/tex] is the probability of both A and B happening.
P(A) is the probability of A happening.
Outcomes for the dice:
For the pair of dice:
(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)
So 36 total outcomes.
In this question:
Event A: Sum of 8
Event B: One dice resulting in 3.
Probability of a sum of 8:
These are the following desired outcomes:
(2,6), (3,5), (4,4), (5,3), (6,2)
5 outcomes out of 36, so:
[tex]P(A) = \frac{5}{36}[/tex]
Probability of a sum of 8 and one dice resulting in 3.
(3,5) or (5,3), so 2 outcomes out of 36, and:
[tex]P(A \cap B) = \frac{2}{36}[/tex]
Probability that one die resulted in a 3, given that the sum is 8:
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{\frac{2}{36}}{\frac{5}{36}} = \frac{2}{5} = 0.4[/tex]
0.4 = 40% probability that one die resulted in a 3.
